find all values for k such that the eq. x^3 −13x+k=0 has three integer roots.


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Question Number 200388 by mr W last updated on 18/Nov/23

$f i n d a l l v a l u e s f o r k s u c h t h a t t h e e q .$ $x^{3} - 13 x + k = 0 h a s t h r e e i n t e g e r r o o t s .$
	Answered by Frix last updated on 18/Nov/23
	$(x−p)(x+(p/2)−q)(x+(p/2)+q)=0 x^3 −((3p^2 +4q^2 )/4)x−((p(p^2 −4q^2 ))/4)=0 ⇒ q=((√(52−3p^2 ))/2)∧k=−p(p^2 −13) ⇒ p=±{1, 3, 4}∧k=±12$
	$(x - p) (x + \frac{p}{2} - q) (x + \frac{p}{2} + q) = 0$ $x^{3} - \frac{3 p^{2} + 4 q^{2}}{4} x - \frac{p (p^{2} - 4 q^{2})}{4} = 0$ $\Rightarrow$ $q = \frac{\sqrt{52 - 3 p^{2}}}{2} \land k = - p (p^{2} - 13)$ $\Rightarrow p = \pm {1, 3, 4} \land k = \pm 12$
	Commented by mr W last updated on 18/Nov/23

	$t h a n k s s i r!$
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